IIT JEE , AIEEE Forum - Mathematics , Algebra - is there any shortcut method 4 solving this:

coolpals_16

any shortcut method 4 this......

if ${(1+x-2{x}^{2})}^{6}$ = $1+ {a}_{1}x +{a}_{2}x^{2}+{a}_{3}{x}^{3}.........................+{a}_{12}{x}^{12}$

then a₂+a₄+a₆+......a₁₂=?

1)32 2)63 3)64) 4)none

coolpals_16

the way i m solving it is 2 write l.h.s as $(-2)^6 {(x-1)^6(x+1/2)^6}$ ,further expaning each of these and then equating coefficients....which makes it very lengthy.....plz suggest shortcut methods,if any

krishna.gopal

First put x=1 and you will get
1+a1+a2...+a12=0
Now put x=-1
1-a1+a2-a3....=a12 = 4
Adding th two we get
a2=a4+a6....a12=4/2=2

coolpals_16

thanx a lot

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